Introduction to Rotational Dynamics
When we looked at objects moving in a circle in Further Mechanics Tutorial 1 we only considered small objects moving around a central point. The objects themselves were small compared to the radius of their orbit. Examples included model aeroplanes tethered by a string, or planets moving around a star. These objects were held to their paths by centripetal force and if that force were stopped, the objects would fly off tangentially in a straight line.
These ideas don't work for rotating systems in which an object is spinning on its axis. Examples of these include:
rotors of electric motors;
flywheels.
You will come across quite a lot of odd looking symbols. Don't worry; they are Greek letters which are used as Physics codes. A lot of the equations are identical in concept to the equations of linear motion (motion in a straight line) which you came across in Mechanics Tutorial 6 (AS).
Let's look at the odd symbols and what they mean:
Symbol 
Pronounced 
Greek Letter 
Code for 
a 
Alpha 
a 
acceleration 
w 
Omega 
long o  ō as in "mōtion" 
angular velocity 
q 
Theta 
"th" as in "therapy" 
angle 
We will now have a look at some of the terms used in rotational dynamics:
Angular velocity is an important quantity. Suppose we had a record deck turntable turning at 33 rpm (revolutions per minute). Clearly an object at the rim of the turntable moves with a higher linear speed that an object towards the centre. However both move through the same angle every second. We call this the angular velocity, which has the physics code w. We don't use degrees per second, but radians per second. Go back to Further Mechanics Tutorial 1 if you are not sure what a radian is (1 rad » 57^{o}). Radians are dimensionless units, so are ignored in unit analysis. Some purists say they should be left out altogether. In these notes I will always include them.
Every revolution, all parts of the object turn through 2p radians.
A common way of expressing the rate of turning is revolutions per minute. All the equations we will use need the rate of turning (angular velocity) to be in radians per second. Therefore be a good chap and convert the revolutions per minute to radians per second.
Angular velocity = 2p(rev min^{1} ÷ 60 s min^{1})
Note that rpm is better written min^{1}. But you won't be penalised for writing 'rpm'.
What is the angular velocity of a hifi record deck playing at 33 rpm? 
Angular displacement is simply the angle turned in any given direction. It is given the code q and is measured in radians.
If the angular velocity is changing, it is of course accelerating. So we have a term angular acceleration, given the code a.
What do you think the units are for angular acceleration? 
So, just like (linear) velocity = displacement ÷ time, we can write:
angular velocity = angular displacement ÷ time
And just like
acceleration (m s^{2}) = change in velocity (m s^{1}) ÷ time interval (s)
we can write
angular acceleration (rad s^{2}) = change in angular velocity (rad s^{1}) ÷ time interval (s)
It takes a motor 5 seconds to accelerate from rest to 3000 rpm. What is its angular acceleration? 
All these are vector quantities, but the directions are pretty easy, clockwise or anticlockwise. No horrible sines or cosines. So far pretty easy, what? Now it gets a bit harder...
You will know that any mass has a reluctance to move, which we call inertia. In linear dynamics, we say that all objects are point masses. We have an analogous situation in rotational dynamics which we call the moment of inertia. The moment of inertia is the measure of the opposition of a rotating body to angular acceleration. It is given the physics code I and its units are kg m^{2}.
Suppose we have a disc radius r spinning with angular velocity w rad s^{1} . We can think of it as made up of lots of little point masses. We know that the linear speed of each point is v, where v = wr. We also know that:
Each little point mass m therefore has a kinetic energy:
The total kinetic energy can be found by adding up all the kinetic energies of the little point masses:
We can rewrite this as:
And we can rewrite this as:
The term Smr^{2} is the sum of all the terms mr^{2}. This term is described as the moment of inertia and is given the code I.
Its units are kilogram metre squared (kg m^{2}). The strange looking symbol S is Sigma, a Greek capital letter 'S'. It means "the sum of".
For the kinetic energy:
In the exam you are NOT expected to then go on to integrate to derive for particular cases of moments of inertia. The appropriate relationship will be given to you, or the moment of inertia will be given to you already worked out. However if you want to see this, click to go to Tutorial 1 B. The derivations are long and are not intuitive.
For a circular disc of mass M and radius r:
For a solid cylinder, the relationship is as above.
For a hollow cylinder open at both ends, the moment of inertia is:
For a solid sphere, mass M and radius r the moment of inertia is:
For a thin spherical shell, mass M and radius r the moment of inertia is:
A circular disc and a solid sphere each has a mass of 2.5 kg and a radius of 0.2 m. What is the moment of inertia for each one? 
In the past papers I have not yet seem the first or third of these formulae, but I have seen a question that got you to use I = Mr^{2}, ignoring the sigma bit. Other times you have been asked to find I from either the torque or the kinetic energy.
Extension (for SQA Advanced Higher Students only)
For a rod of length l, and mass m, the moment of inertia can be worked out using the relationships below. If you want to see the derivation, click HERE.
If the rod is spinning about its centre point like this:
The relationship is:
If the rod is held at one end like this:
The relationship is:
The key thing to remember is that the moment of inertia is the rotational equivalent of mass.
The kinetic energy of a spinning object is given by:
[I  moment of inertia (kg m^{2}); w  angular velocity (rad s^{1})]
Like kinetic energy in linear motion, rotational kinetic energy is a scalar, even though the angular velocity is a vector.
A flywheel battery can be used in place of leadacid batteries to provide a shortterm electrical power supply when mains power fails. Energy is stored as rotational kinetic energy in a rapidly spinning rotor, which is driven up to speed by a mainspowered motor. To recover the energy, the motor is operated in reverse as a generator driven by the spinning rotor.
The rotor of a flywheel battery is a thinwalled circular cylinder of mass 160 kg and mean radius 0.34 m, which can be rotated at a maximum safe angular speed of 44 000 rev min^{1}.
Calculate: (a) the moment of inertia of the rotor about its own axis,
(b) the rotational kinetic energy stored in the rotating rotor when it spins at 44 000 rev min^{1}. (AQA past question) 
To show you that this is not a figment of the examiner's imagination, the picture below shows how a flywheel battery works
Image by Tosaka, Wikimedia Commons
(1) Case; (2) Flywheel; (3) Motorgenerator; (4) Main bearing; (5) Transformer; (6) Vacuum pump; (7) Current in; (8) Current out.
This idea is not new. Thirdrail electric locomotives had a motorgenerator to help them trundle over gaps in the electric conductor rail.
The total potential energy of a rotating system that is in linear motion is the sum of:
the translational kinetic energy;
the rotational kinetic energy.
Worked Example A solid ball has a mass of 1.50 kg and a diameter of 10.0 cm. It is spinning at 2000 revolutions per minute and is travelling horizontally at a linear speed of 20.0 m s^{1}.
Calculate: (a) The moment of inertia; (b) The rotational kinetic energy; (c) The total potential energy. 
Answer Radius = 0.050 m. Rate of rotation per second = 33.3 s^{1}. Angular velocity = 33.3 × 2 p = 209.4 rad s^{1}.
(a) Formula for the moment of inertia of a sphere:
I = (2 × 1.50 kg × (0.050 m)^{2}) ÷ 5 = 1.50 × 10^{3} kg m^{2}.
(b) Rotational kinetic energy:
E_{k} = 1/2 × 1.50 × 10^{3} kg m^{2} × (209.4 rad s^{1})^{2} = 32.9 J
(c) Translational kinetic energy: E_{k} = 1/2 × 1.50 kg × (20.0 m s^{1})^{2} = 300 J
Potential energy = translational kinetic energy + rotational kinetic energy
= 300 J + 32.9 J = 333 J (3 s.f) 
Many machines have a flywheel which is a heavy lump of metal spinning on an axis. A single cylinder petrol engine needs a flywheel to keep it running smoothly. If it didn't have one, it would stall on the compression stroke. Fourcylinder car engines would still be jerky without a flywheel. Twelve cylinder car engines have less need for a flywheel, but still have one because:
It is useful to put the toothed ring for the starter motor onto it;
It is a useful face for a clutch.
What is the best design for a flywheel? The obvious shape is a solid disk of steel. The diagram shows a couple of crosssections. Both flywheels have the same mass and mean radius. Both have a small flange of very small mass so that they can be bolted to the shaft.
A solid disk flywheel:
A flywheel that has a thin plate in the middle and most of its mass as a ring around the outside:
Question 6 (Harder) 
By considering the amount of kinetic energy each one can store, which one is the better design? State what assumptions you make. 
The flywheel battery in question 5 would have a flywheel of the second design.
Equations of Rotational Motion
There are four of these, each of which has an equivalent of linear motion, and are used in exactly the same way. The only difference is the code for some of the terms, and if you are at ease with angular velocity, angular displacement, and angular acceleration, you will have no difficulty at all with these. If you have forgotten about the linear equations of motion, go back to Module 2 and revise them.
This table shows the terms and what they mean:
Term 
What it means 
Units 
w_{1} 
Angular velocity at start 
rad s^{1} 
w_{2} 
Angular velocity at end 
rad s^{1} 
q 
Angular displacement 
rad 
a 
Angular acceleration 
rad s^{2} 
t 
time 
s 
(1) Used to link a second angular velocity to the first, acceleration and time.
It is used just like v = u + at
A car wheel is being tested for balance. It is spinning at 750 rpm and the machine then accelerates it to 1500 rpm over a period of 3 s. What is the angular acceleration? 
(2) Used to link the angular displacement to the angular velocity at the start , with the time, and the angular acceleration.
It is used just like s = ut + 1/2 at^{2}.
Use your result from question 7 to work out the angular displacement through which the wheel turns as it accelerates from 750 rpm to 1500 rpm. 
(3) Used to link the angular velocity at end with the first angular velocity , the angular displacement and the acceleration.
It is used just like v^{2} = u^{2} + 2as.
In a recent experiment to test a "crashproof" fuel, an aeroplane was deliberately crashlanded onto spikes which were designed to slice the fuel tanks open. The test was a failure because the fuel burst into flames. The reason for this was because one of the engines was stopped by a spike. The turbine was running at 30 000 rpm and was stopped within 1/3 revolution. The kinetic energy of the engine was dissipated as intense heat which acted as an ignition source for the fuel.
What was the angular deceleration suffered by that engine? 
(4) This equation is used to link the angular displacement to the average of two angular velocities and the time interval:
It is used just like s = (u + v)t/2
A car wheel is being tested for balance. It is spinning at 750 rpm and the machine then accelerates it to 1500 rpm over a period of 3 s. Use the above equation to work out the angular displacement. How does your answer compare to the answer you worked out in Question 8? 
Torque and Angular Acceleration
You will be familiar with F = ma. Rotational motion has an identical relationship. You will remember that a torque is a turning force.
The diagram shows a cord wrapped around a pulley (like the pullcord used to start a lawn mower). The pulling force F causes a torque t ( tau, a Greek lower case letter 't', the physics code for torque) which can easily be worked out by:
t = Fr
If we double the pull, we will double the torque which means that the angular acceleration will be doubled. However if we look at the mass, we find that it's not the only factor. The shape of the pulley is important, so instead of mass, we use the moment of inertia.
So if we apply a torque to a rotating body, it will undergo angular acceleration. In other words it will spin faster. The angular acceleration due to a torque is given by:
t = Ia
[ t  torque (N m); I  moment of inertia (kg m^{2}); a  angular acceleration ( rad s^{2})]
The wheel of a large dumper truck has a moment of inertia of 10 000 kg m^{2} and is being tested by being rotated at 60 rpm. It is brought to rest in a time of 40 s. (a) What is the initial angular velocity in rad s^{1}? (b) What is the angular acceleration? (c) What is the angular displacement in the first 20 s? (d) What is the applied torque? 
Let's sum up by comparing linear motion with rotational motion:
Linear Motion 
Rotational Motion 

Quantity 
Code 
Unit 
Quantity 
Code 
Unit 
Displacement 
s 
m 
Angular displacement 
q 
rad 
Velocity 
v 
m s^{1} 
Angular velocity 
w 
rad s^{1} 
Acceleration 
a 
m s^{2} 
Angular Acceleration 
a 
rad s^{2} 
Mass 
m 
kg 
Moment of Inertia 
I 
kg m^{2} 
Force 
F 
N 
Torque 
t 
N m 
We will compare equations:
Linear Motion 
Rotational Motion 




F = ma 
t = Ia 
v = u + at 







Using Calculus in Rotational Dynamics (Extension)
In linear dynamics (Mechanics Tutorial 6) we saw how we can use calculus to establish relationships between quantities like acceleration, velocity, and displacement. We can do the same in rotational dynamics. We do not get sudden transitions from angular acceleration to constant angular velocity, with sudden graphical inflections. At university level, a calculus treatment is preferred. Calculus is a mathematical technique that allows us to work out the gradient of a graph or the area under the graph, if we know the relationship between two quantities.
In rotational dynamics you will see the following:
1. Angular velocity, w, is the rate of change of angular displacement q. In calculus notation, this is written as:
2. Angular acceleration, a, is the rate of change of angular velocity, w. In calculus notation, this is written as:
So what is the difference? If we are talking about a constant rate of change, there is no difference. However, if the rate of change is variable, the calculus notation is used for an instantaneous change. This is summed up in the graph below:
In this graph we see a very irregular increase in the angular velocity of an object. We can look at the rate of change in velocity at a particular instant, or we look at the overall change in velocity over a longer period. The instantaneous change in the angular velocity is represented by the term dw, while the overall change is represented by Dw.
We could, of course, take a tangent from the graph at the particular instant and measure the gradient of the tangent. However it is likely that there will be uncertainty. With differentiation, there is no uncertainty.
Acceleration is related to displacement using:
This is a second derivative, which means a derivative of a derivative.
In summary:
Using Differentiation
Differentiation is about finding the gradient of the graph. As we have seen above:
Velocity is the gradient of the displacement time graph;
Acceleration is the gradient of the velocity time graph
Maths Window There are two things that we can do with Calculus: 1. We can differentiate, which means that we find the gradient of the graph of a known relationship. 2. We can integrate, which means that we find the area under the graph of a known relationship.
We do these mathematically without having to draw the graph.
This is NOT a comprehensive treatment of calculus, but I hope it will help you how to use it in kinematics calculations
Differentiation Differentiation is about determining the gradient of a graph.
There are a number of rules of differentiation We will use only two here.
Let us suppose we have a a straight line graph that follows the general relationship:
y = mx + c If we want to differentiate this, we get:
Therefore:
This tells us that the gradient is m. 
We can use calculus to work out the velocity at a particular instant. Here is a graph showing the displacement of an object subject to constant acceleration. This graph has been drawn using the equation:
The value of the acceleration is 4.0 rad s^{2}.
We could, of course, work the gradient of the tangent to give us the velocity at exactly 6.0 s, but there will be uncertainty. Instead, let's use calculus:
Therefore we can substitute:
w = 4.0 rad s^{2} × 6.0 s = 24 rad s^{1}
This is a lot easier that drawing a tangent, then measuring the rise and the run.
We can differentiate the equation:
As below:
To give us:
Using Integration
Integration is about finding the area under the graph. In Mechanics 6 we have seen how the kinematics equations have been worked out using the area under the velocity time graph. the same applies to rotational dynamics. For some questions we have simply counted the squares under the graph. The counting of squares is both tedious and prone to uncertainty. In simple graphical treatments, acceleration is counted as constant. In reality it is not. Therefore a mathematical approach is more satisfactory, as it can be quicker and is less prone to uncertainty.
Consider a rotor of a motor accelerating at a rate of a rad s^{2} over a period of t s. This is a real world situation, and the acceleration is not constant, but reduces because of the increase in friction from the road and the air resistance.
Suppose we want to find the angular displacement q between the points A and B. We know that angle turned is the area under the graph. We could use the equation:
The equation works out the area under the orange line in the graph. The answer it would give would be too low.
Alternatively, we could break the area into three little strips as shown in the graph. So we could work out the area of each little strip and add them together:
This will give us a better answer, but it will only be an approximation. We could make the strips narrower, using a shorter time interval.
Therefore the answer gets closer to the real answer, but it is still an approximation. However, if we make the time interval infinitesimally small, we end up with the true answer. This we do by the process of integration. Instead of writing the width of each strip a Dt, we write dt. Integration adds up all the little strips to give us the true answer.
Maths Window Integration Integration is the reverse process of differentiation. The idea is shown in the picture below:
The 2x term is the function. The function to be integrated is sometimes called the integrand. Therefore we write this in calculus form as:
The dx term shows that the little strips go along the xaxis. The integration symbol is a fancy capital letter 'S', which means "summed together". So we now write:
The C term is a constant. When we differentiated, the constant that was added to the function had a differentiated value of 0. Now we are applying the process in reverse, we need to have a definite value for the constant.
Here are some rules for integration:
Often you will need to integrate between two points. You may see an equation like this:
This means you have to work out the value of the integral at p and the value of the integral at q, and then subtract one from the other. This is shown in the picture below. The constant, C = 0 for the sake of this argument.

Consider an object moving with a constant angular acceleration, a rad s^{2} from an initial angular velocity at t = 0, w_{1} rad s^{1}, to a final angular velocity, w_{2} rad s^{1} at time t. This is shown on the graph below:
We know that this graph shows the equation:
We also know that the area under the graph is the displacement. So we can can integrate, since we know that:
So we can write:
Therefore:
So let's use that result:
Worked Example An object has an initial angular velocity of 6.5 rad s^{1}. At time t = 0 it accelerates at a rate of change of angular velocity of 2.85 rad s^{1} . What is the angle turned between time t = 3.6 s and time t = 8.5 s? 
Answer We can write this in calculus notation:
Therefore:
